p$. Then $u\in L^{(\alpha-p)/p}(0,R)$ for any solution $u$
of (\ref{meqn}).
\end{thm}
In Section 3, we also show that for $H(r,s)=g(r)f(s)$, the
following result holds.
\begin{thm} \label{FKO}
Let $H(r,s)=g(r)f(s)$ satisfy (H1)--(H3),
with $g(0)>0$ and $g$ positive, non-decreasing near $R$. Suppose
(\ref{meqn}) has a blow-up solution $u$ such that $u\in
L^{(\alpha-p)/p}(0,R)$ for some $\alpha>p$. If $g\in L^{1/\sigma}(0,R)$ with
$0<\sigma0$
imply that $f(s)>0$ for $s>0$. Since $f(s)>0$, it follows from
(H1) that $g$ is non-negative on $[0,R)$.
\end{rem}
Finally, we give some corollaries to Theorem \ref{FKO}.
\section{A Comparison Result}
We will need the following comparison lemma (see \cite{RAW} for a
proof). For notational convenience in stating the lemma and in
this section, we let $L$ denote the differential operator on the
left hand side of equation (\ref{meqnre}) above. In this lemma,
we use the following notation: $u(a+)0$ such that $u1$, then
$$
\lim_{t\to\infty}\frac{t^\alpha}{F(t)}=0.
$$
\end{lem}
We need the following lemma, which shows that solutions
of (\ref{meqn}) with initial slope zero have non-decreasing slope
for $r \in [0,R)$.
\begin{lem}\label{prel}
Suppose in addition to (H1)--(H3), $H(r,\cdot)$ is
non-decreasing on $[0,R)$. If for $0w_k(r)$ for some $0 1/R$.
For $t,r\in (0,R-1/k)$, and $n>k$ we have
\[
|w_n(r)-w_n(t)|=\big|\int_t^rw'_n(s)\,ds\big| \leq
w'_n(\zeta)|r-t| \leq w'_{k+1}(R-1/k)|r-t|,
\]
where $\zeta=\max\{r,t\}$. The fact that $w'_{k+1}$ is
non-decreasing, by Lemma \ref{prel}, has been exploited in the
last inequality.
Thus $\{w_n\}_{n=k+1}^\infty$ is a bounded equicontinuous family
in $C([0,R-1/k])$, and hence has a uniformly convergent
subsequence. Let $w$ be the limit. For $r\in[0,R-1/k]$ and $n>k$
the solution $w_n$ satisfies the integral equation
$$
w_n(r)=w_n(0)+\int_0^r\Big(\int_0^t\left(\frac{s}{t}\right)^\lambda
H(0,w_n(s))\,ds\Big)^{1/(p-1)}\,dt\,.
$$
Letting $n\to\infty$
we see that $w$ satisfies the same integral equation. Since $k$ is
arbitrary we conclude that $w$ satisfies equation (\ref{seqn}).
Since $u\leq w_n$ on $(0,R-1/k)$ for each $n\geq k$ we conclude
that $u\leq w$ on $(0,R)$.
\end{proof}
\section{Proofs of Main Results and Some Corollaries}
\begin{proof}[Proof of Theorem \ref{uint}]
By Theorem \ref{scomp} we take a solution $w$ of
(\ref{seqn}) such that $u(r)\leq w(r)$ for $0\leq r\frac{1}{\lambda+1}f(w),\quad 00$. Suppose
(\ref{meqn}) has a blow-up solution that belongs to
$L^{(\alpha-p)/p}(0,R)$ for some $\alpha>p$. If $g$ is bounded, then
$F\in KO(\gamma)$ for any $0<\gamma<\alpha$.
\end{cor}
\begin{rem} \label{rmk3.2}\rm
The conclusion of Corollary \ref{coro3.1} is false when $g$ is
unbounded near $R$ as the following example shows.
The function $u(r)=(1-r)^{-1}$ is a solution of
\begin{gather*}
u''(r)=g(r)f(u),\\
u(0)\geq 0,\quad u'(0)\geq 0,\quad u(1)=\infty,
\end{gather*}
where
$$
g(r):=2/(1-r),\quad \text{and}\quad f(s):=s^2.
$$
Observe that
$u\in L^{(\alpha-2)/2}(0,1)$ for $2<\alpha<4$. However note that
$F\notin KO(3)$.
\end{rem}
\begin{cor} \label{coro3.2}
Suppose $H(r,s)=g(r)f(s)$ satisfies (H1)--(H3), with $g(0)>0$, $g$
non-decreasing on $[0,R)$, and $f(0)=0$. Further, let $g$ be
bounded on $[0,R)$, and let $F\in KO(p)$. Then a blow up solution
$u$ of (\ref{meqnre}) belongs to $L^q(0,R)$ for some $q>0$ if and
only if $F\in KO(\gamma)$ for some $\gamma>p$.
\end{cor}
\begin{proof}
Suppose $F\in KO(\gamma)$ for some $\gamma>p$. Then by Theorem
\ref{uint}, we see that $u\in L^q(0,R)$ for $q=(\gamma-p)/p$. For
the converse, suppose that $u\in L^q(0,R)$ for some $q>0$. Then
for $\alpha=p(q+1)$ we see that $q=(\alpha-p)/p$ so that by the
above corollary, $F\in KO(\gamma)$ for some $p<\gamma